PProblem 10.1

Let A1 and and web respectivelsions give, the total weight density of steparallel axis theorem

I

I

I

d

I

W

AA

to

Problem 10.2

From = MySo we have

As you canthe yield strength anclearly not good andcylinder that can be ureason and if it is stothe yield stress. In th( )( )psi10120psi10151

120625.01029

102910200

336

26

29

>=

=

==

y

steel inlbs

m

NE

roblem Set #10 Solution 1.050 Solid Mechanics Fall 2004

0.571 in A1

A2 be the cross-sectional areas of the ange y. Then we can compute, with the dimenarea, the weight per unit length (given the d1 el) and the moment(s) of inertia - using the 18.006 in . A2

0.355 in

7.486 in

( )( ) ( )( )[ ]( )

( )( ) ( ) ( )( )( ) ( ) ( )( ) ( )( )

( )( )[ ]( ) ( ) ( )( )

4

33

4

1

2 1

33

33

2 121

2 2

2 1

12 2

12

2

12 2

12

49.45

7850

4.275

in

in

in

d

ft lbs

in lbs

L

in lbs

m

kg inAAA

in in

yy

yy

xx

xx

steel

=

+

=

=

==

++

=

===

==

=++=

==

==

99.39

486.7 571.0 355. 0 571.0 2 006.18

87.791

7175.8 0.571 -18.006

275.4 571.0 486.7 571.0 2 006.18 355.0

1212.4 2835.0 537.14

2835.0

537.14 987. 5 355.0 571.0 2 006.18

571.0 486.7

A tal

by/I and Mb = EI/, we get that y =Ey/.

( )( ) psi10120psi10151

12 1029

102910200

33 6

2 6

2 9

>=

=

==

y

steel in lbs

m

NE

0625.0

see, if the radius of cylinder is reduced to 12 in, the stress in the steel wire will exceed d thus make the steel wire fail (or undergone permanent plastic deformation). This is thus you should tell your boss about this and inform him that the minimum radius of the sed for this steel wire storage is 16 in (which will give y = 113x103 psi). But for safety

red for a long time period, you might want to keep the stress in the steel wire well below is case, the original 20 in radius is a good choice.

-Problem 10.3

A steel reinfo

d

Tota

b

h neutral axis

a range of realistic

Make a sketch ofthe diameter of the ro

The magnitmaximum tensile str

If the steel and concrmust be 10 times theYoungs moduli give

d--- = (1.2h

We plot this linear repositive and both les

We can gentionship between theallow for different ra

Letting C

dwe obtain --- =h

The intercethrough the point (1,for the particular cas

Now we joidirection of the axis

dhWe can eliminate be-- ---

--- ---

rced beam is to be made such that the steel and the concrete fail simultaneously.

c

s

Mb

y

l area = n*As

h

y

If Es = 30 e06 psi steel

Ec= 3.6 e06 psi concrete and taking

failure steel = 40,000 psi failure concrete = 4,000 psi (compression)

how must be related to d/h for this to be the case?

Now, letting 2 Es nAs Ec bh

--------------------------= find d/h and values for values for the area ratio, (nAs/bh), hence for a range of values for .

one possible composite cross-section showing the location of the reinforcing rod. Take d as 0.5 inches.

ude of the maximum compressive stress in the concrete will be seen at y = (h-d). The ess in the steel will be seen at y = -(d-h). They are1:

Mb (h d)E Mb (d h)Ec s = ----------------------------------- and = --------------------------------------c smax maxEI EI

ete are to fail at the same bending moment, then the maximum tensile stress in the steel maximum compressive stress in the concrete. So (d h)E = 10 (h d )E . With thes c n this can be written,

d)1 --- or = 2.2 ---d 1.2h h 1.0lationship at the right. Beta and d/h must be s than 1.0

eralize this relationship if we allow the rela- maximum stresses to be arbitrary and also tios for the Youngs moduli.

Es max c= ----------------- ------

c max Es 0 .546.25 .75 1.0 d/h

C 1 --- + dh or = (1 C) ---d

C h

pt for =0 is then just d/h = C/(1+C). Furthermore, all lines for any C must pass 1). We show a shaded region which would include all plots for (1/3)

This, in time, simpli

or, factoring the quad

We can not have d/h

To answer the questiof steel area to bea

then

We set out in a table

We see the percent a10 x 15 inches, and tter of 0.875 inches, (With this xsection, th

1. Cody: This is why2. .5 in diameter rodes considerably: d d --- --- = 2

2 + 1 + d --- 1C 0 hh h ratic in d/h

d =

--- 1 2 + d --- 1C 0h h

= 1, so this common term may be factored out1 and we must then have d= 1 --- Ch

2E nA nAs nAs s son posed, if C = 1.2 = ------------------ = 16.7--------- and if we let = --------- be the ratioE bh bh bhcm xsection area,

d d h ( )16.7= 1 --- 1.2 or = 1 --- 20h

, some realistic values for d/h, , the area fraction of steel, and nally .

d/h 0.55 0.022 0.01 0.575 0.021 0.065 0.6 0.020 0.12 0.625 0.018 0.175 0.65 0.017 0.23 0.675 0.016 0.285

rea of steel is about 2% in all cases. If we choose, arbitrarily, a beam cross section of say ake d/h =0.6, the area of the steel will be .020x150in2 = 3 in2. If the rods have a diamereinforcing bar #7)2 each then has an area of 0.60 inches2, so we will need 5 rods. e steel will fail in tension, the concrete in compression, at the same time.

d/h =.6

= .12 h = 1.8 in.

~1.5

1.5in cover for re protection

= 0.02

10 in.

15 in

you always found the root of the quadratic to be d/h = 1.0. s will not t as well as these with diameter 0.875.